Question

How do you solve `[(-x,+2y,-4z,-5u,=,27),(-2x,-3y,+3z,-3u,=,9),(5x,-4y,+2z,-3u,=,5), (-2x,+0y,+4z,-5u,=,29)]`?

Answer 1

`x=0`, `y=3`, `z=1` and `u=-5`

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

`A=((-1,2,-4,-5,|,27),(-2,-3,3,-3,|,9),(5,-4,2,-3,|,5),(-2, 0, 4, -5,|,29))`

I have written the equations not in the sequence as in the question in order to get `1` as pivot.

Perform the folowing operations on the rows of the matrix

`R2larrR2-2R1`, `R3larrR3+5R1, R4larrR4-2R1`;

`A=((-1,2,-4,-5,|,27),(0,-7,11,7,|,-45),(0,6,-18,-28,|,140),(0, -4, 12, 5,|,-25))`

`R2larrR2-2R4`;

`A=((-1,2,-4,-5,|,27),(0,1,-13,-3,|,5),(0,6,-18,-28,|,140),(0, -4, 12, 5,|,-25))`

`R1larrR1-2R2`, `R3larrR3-6R2, R4larrR4+4R2`;

`A=((-1,0,22,1,|,17),(0,1,-13,-3,|,5),(0,0,60,-10,|,110),(0, 0, -40, -7,|,-5))`

`R1larrR1`, `R3larrR3*2`, `R4larrR4*3`;

`A=((1,0,-22,-1,|,-17),(0,1,-13,-3,|,5),(0,0,120,-20,|,220),(0, 0, -120, -21,|,-15))`

`R4larrR4+R3`;

`A=((1,0,-22,-1,|,-17),(0,1,-13,-3,|,5),(0,0,120,-20,|,220),(0, 0, 0, -41,|,205))`

`R4larr(R4)/(-41)`;

`A=((1,0,-22,-1,|,-17),(0,1,-13,-3,|,5),(0,0,120,-20,|,220),(0, 0, 0, 1,|,-5))`

`R1larrR1+R4`, `R2larrR2+3R4`, `R3larrR3+20R4`;

`A=((1,0,-22,0,|,-22),(0,1,-13,0,|,-10),(0,0,120,0,|,120),(0, 0, 0, 1,|,-5))`

`R3larr(R3)/120`;

`A=((1,0,-22,0,|,-22),(0,1,-13,0,|,-10),(0,0,1,0,|,1),(0, 0, 0, 1,|,-5))`

`R1larrR1+22R3`, `R2larrR2+13R3`;

`A=((1,0,0,0,|,0),(0,1,0,0,|,3),(0,0,1,0,|,1),(0, 0, 0, 1,|,-5))`

Thus `x=0`, `y=3`, `z=1` and `u=-5`